
For a given material, the Young’s modulus is 2.4 times its modulus of rigidity. What is the value of Poisson's ratio?
${\text{A}}{\text{.}}$ 0.2
${\text{B}}{\text{.}}$ 0.4
${\text{C}}{\text{.}}$ 1.2
${\text{D}}{\text{.}}$ 2.4
Answer
610.8k+ views
Hint- Here, we will proceed by obtaining a relation between Young’s modulus of elasticity and modulus of rigidity according to the problem statement and then, substitute this relation in the formula which represents the relation between Young’s modulus of elasticity, Bulk modulus (modulus of rigidity) and Poisson’s ratio.
Complete step-by-step solution -
Formula Used- E = 2G(1+$\nu $).
Let the young’s modulus and modulus of rigidity for the given material be E and G respectively
It is given that the Young’s modulus is 2.4 times its modulus of rigidity for the given material
i.e., E = 2.4(G)
Also, assume the Poisson’s ratio for the given material be $\nu $
According to the relation between the Young’s modulus of elasticity, Bulk modulus (modulus of rigidity) and Poisson’s ratio, we have
Young’s modulus of elasticity is equal to twice the product of modulus of rigidity and one plus Poisson’s ratio
i.e., E = 2G(1+$\nu $) $ \to (1)$ where E denotes the Young’s modulus of elasticity, G denotes modulus of rigidity (or bulk modulus) and $\nu $ denotes the Poisson’s ratio
By substituting E = 2.4(G) in equation (1), we get
$ \Rightarrow 2.4\left( {\text{G}} \right) = 2{\text{G}}\left( {1 + \nu } \right)$
Modulus of rigidity (G) with cancel out from both the sides in the above equation, we will get
$
\Rightarrow 2.4 = 2\left( {1 + \nu } \right) \\
\Rightarrow 1 + \nu = \dfrac{{2.4}}{2} \\
\Rightarrow 1 + \nu = 1.2 \\
\Rightarrow \nu = 1.2 - 1 \\
\Rightarrow \nu = 0.2 \\
$
Therefore, the Poisson’s ratio for the given material is equal to 0.2.
Hence, option A is correct.
Note- Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force. Tensile deformation is considered positive and compressive deformation is considered negative. For most of the materials, the value of Poisson’s ratios lies from 0 to 0.5.
Complete step-by-step solution -
Formula Used- E = 2G(1+$\nu $).
Let the young’s modulus and modulus of rigidity for the given material be E and G respectively
It is given that the Young’s modulus is 2.4 times its modulus of rigidity for the given material
i.e., E = 2.4(G)
Also, assume the Poisson’s ratio for the given material be $\nu $
According to the relation between the Young’s modulus of elasticity, Bulk modulus (modulus of rigidity) and Poisson’s ratio, we have
Young’s modulus of elasticity is equal to twice the product of modulus of rigidity and one plus Poisson’s ratio
i.e., E = 2G(1+$\nu $) $ \to (1)$ where E denotes the Young’s modulus of elasticity, G denotes modulus of rigidity (or bulk modulus) and $\nu $ denotes the Poisson’s ratio
By substituting E = 2.4(G) in equation (1), we get
$ \Rightarrow 2.4\left( {\text{G}} \right) = 2{\text{G}}\left( {1 + \nu } \right)$
Modulus of rigidity (G) with cancel out from both the sides in the above equation, we will get
$
\Rightarrow 2.4 = 2\left( {1 + \nu } \right) \\
\Rightarrow 1 + \nu = \dfrac{{2.4}}{2} \\
\Rightarrow 1 + \nu = 1.2 \\
\Rightarrow \nu = 1.2 - 1 \\
\Rightarrow \nu = 0.2 \\
$
Therefore, the Poisson’s ratio for the given material is equal to 0.2.
Hence, option A is correct.
Note- Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force. Tensile deformation is considered positive and compressive deformation is considered negative. For most of the materials, the value of Poisson’s ratios lies from 0 to 0.5.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

